Structural Engineering & Statics
where the load goes
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The load must go somewhere — page 7

Essays 145 to 156 of 156 on this thread, in the same order.
The tendon is a load, pointing the other way. A 14 m beam with a parabolic tendon dropping 260 mm to midspan, stressed to 1440 kN after losses. Its curvature pushes the beam up along its whole length with an intensity of 8Pe/L² = 15.28 kN/m, against an applied 17.63 kN/m — so 2.34 kN/m is left to bend anything, and the beam carries 57.4 kNm where an unstressed one carries 432 kNm. What the section then feels is 6.40 MPa of uniform compression and very little else. Internal forces

The load that comes from changing direction

A force that travels in a straight line asks nothing of anything. Bend its path and it asks for a transverse load of F over R along every millimetre of the curve, and that load is real, is nowhere on the load schedule, and is the same statement behind a prestressing tendon, a hoop force, an arch thrust and a web that buckles with nothing applied to it.

 A section has two areas and the tables give one of them. Peak shear stress divided by the mean, for four sections of exactly the same gross area and depth. The mean is V/A and is the number a first calculation uses; the peak is what the material actually sees, and the ratio between them is a property of shape alone. A rectangle's is 1.5 — the parabola's peak over its average — and it is one of the few numbers in this subject that is exactly derivable and universally ignored. An I-section's is near 1.98, and the reason is on the second bar: 97% of the shear is inside a web that is 56% of the area. So the flanges carry the moment and almost none of the shear, and the web carries the shear and almost none of the moment — which is why a shear check on an I-section uses the web area and a moment check uses the whole section, and why the two checks are about two different pieces of steel. Sections and stress

The section has two areas

A shear force divided by the area of the section is not the shear stress anywhere in it. A rectangle's peak is exactly one and a half times that number and an I-section's web carries nearly all of the shear over a fifth of the area, which is why a moment check and a shear check on the same member are checks on two different pieces of steel.

 Two curvatures of opposite sign, which is what makes it a structure. A cable net over a 36 m square, drawn as the two families of cables that are also the two rulings of the surface. One family sags and carries downward load by hanging; the other rises and carries upward load — wind uplift, and a load reversal anywhere — by the same mechanism upside down. Neither can do anything alone. A single family of cables is a mechanism: it changes shape freely under any load pattern it was not tensioned for, and the shape it moves to is decided by the load rather than by the designer. Put the two together and each is the other's restraint, but only if they are pulled against one another first — the pretension of 520 kN in the sagging family and 715 in the hogging one is a self-equilibrating state that exists with no load on the roof at all, and it is what turns two mechanisms into one structure. The curvatures are drawn four times their true value: a real net of this span sags 2.2 m over 36, which is flatter than it looks anywhere. Structural form

Two curvatures of opposite sign

A single family of cables is not a structure. It is a mechanism that takes whatever shape the load asks for, and it will do that under any load pattern it was not tensioned for. Cross it with a second family curved the other way, pull the two against each other, and the pair becomes stiff — with no bending anywhere and no material property involved in the stiffness at all.

 Two drawings of one deck, and they are not the same structure. A 112 m viaduct on five supports, articulated two ways. Above, the fixed point is at the left abutment: the far end has to be given 45 mm of movement, and the friction of every sliding bearing runs one way, so the fixed support takes 660 kN before any wind or braking is applied. Below, the fixed point is at the middle pier: the largest joint halves to 22 mm and the friction now cancels across the fixed point, leaving 0 kN. The movement arrows are drawn at 900 times the scale of the deck, because a 45 mm movement on a 112 m span is thinner than the line the deck is drawn with. Nothing about the deck, the loads or the ground has changed between the two. Structural form

Where the structure is allowed to move

One drawing decides how big every movement joint on a bridge is and where every horizontal force goes, it takes an afternoon, and it appears on no calculation sheet. Move the fixed point from an abutment to the middle pier and the largest joint halves and the horizontal force on that support drops from the whole of the friction to none of it.

 The two braces balance until one of them buckles. An inverted-V brace after the compression member has gone. While both braces are elastic they carry equal and opposite forces and their vertical components cancel on the beam above, which is why the beam in a chevron bay is usually sized for gravity alone. The compression brace buckles at 527 kN and then sheds most of what it was carrying — 30% is left here — while the tension brace goes on to yield at 1207. The difference between the two vertical components is 742 kN, applied at the middle of the span with no help from either brace, and it asks the beam for 1483 kNm against the 200 kNm the gravity load asks for — 7.4 times as much. The beam drawn does not: 1683 kNm against a capacity of 731. The force is not a load case anybody applies; it is what the frame leaves behind on its way to the state it will actually be in. Structural form

The force the brace leaves behind

Two braces meeting under a beam carry the storey shear as a tension and a compression whose vertical components cancel, so the beam above sees nothing. They cancel only while both braces are elastic. Once the compression brace buckles it sheds most of its force, the tension brace goes on to yield, and the difference is a point load at midspan that nobody applied.

 Effective length is a property of the storey. The effective length factor of the one column that resists sway, against the total gravity load on the storey as a multiple of its own. At the left-hand end it carries the storey alone and its K is 1.99 — the 2.0 every chart gives a column fixed at the base and free to sway at the top, reproduced here by a route that never mentions a chart. Then columns are added that have pinned bases and therefore no lateral stiffness whatever. They contribute load and nothing else, so they cannot buckle on their own and they lower the load at which everything buckles together. K rises as the square root of the load ratio, exactly, and at the storey drawn — three leaning columns carrying 69% of the gravity load — it is 3.57. That is off the end of every published alignment chart, and the leaning columns themselves, which a designer would take at K = 1.0 for pinned ends, are at 2.54. Stability

The column that leans on its neighbours

A column with a pinned base and a pinned top has no lateral stiffness at all and cannot stand up alone, and yet thousands of them do. What holds them is the rest of the storey, and what it costs is paid by whichever columns do have stiffness — whose effective length rises as the square root of the load being leaned on them, straight off the end of every chart.

 A transverse load with nothing applied. Web slenderness against web thickness, with the limit the flange's own curvature sets. A flange carrying 6213 kN and curved to a radius of 592 m needs 10.5 N per millimetre of radial force to stay on its curve, and the only thing available to supply it is the web. Nothing has been applied to the girder: the load comes from the deflected shape, which is why a straight beam has none of it and a beam at a plastic hinge has a great deal. Setting the radial force against the web's own plate-buckling resistance gives, in four lines, h_w/t_w ≤ k·(E/f_yf)·√(A_w/A_fc) — the form the codes use, arrived at without them. The constants differ: an elastic flange strain gives k = 1.34 and the rule uses 0.3, a factor of 4.5, and the gap is the curvature assumed. k goes as the inverse square root of the flange strain, so 0.3 is a flange strained to 3.4% — which is what a plastic hinge does to it. The rule is not conservative; it is written about a different beam. Stability

The web that is crushed from inside

A plate girder's compression flange is curved by the beam's own deflection, and a curved force needs a transverse load to stay on its curve. The only thing available to supply it is the web. So a deep girder can buckle its web vertically with nothing applied to it at all, and the rule that prevents it is the only clause in the codes about a load no load case contains.

 A bearing capacity is a mechanism, and here it is. Prandtl's collapse mechanism under a 3.0 m footing in a soil of 32° friction. A rigid wedge is driven down with the footing at 61° to the horizontal; a fan of radial shear turns the stress through exactly ninety degrees on a logarithmic spiral whose growth rate is tanφ; and a passive wedge at 29° has to be pushed up and out of the way. Nothing here is empirical — every angle is a function of φ alone — and the mechanism reaches 15.9 m from the centre, which is 10.6 times the footing's half width. That is why two footings closer together than about four widths do not have separate bearing capacities. Equilibrium

The ground is a mechanism

Bearing capacity is met as a formula with three terms and a table of coefficients, and that presentation hides what it is. Underneath is a plastic collapse mechanism — a rigid wedge, a fan of radial shear on a logarithmic spiral, and a passive wedge that has to be pushed up and out of the way — and every coefficient in the table is a property of that one drawing.

 The check that everything adds up, and the error it cannot see. Four versions of the same 3-bay, 4-storey frame, with the global equilibrium residual each one produces — the sum of the reactions against the sum of the applied loads, as a fraction of the applied total. It is the first thing every analysis prints and it is worth having: a lost restraint and a load entered in the wrong unit both show up immediately, at 8% and 32%, because both change what the structure is carrying. The fourth bar is the point. A member whose stiffness is wrong by a factor of ten redistributes the internal forces completely — the second bar shows the change in the member forces, 24% — and the global residual is exactly zero, because the wrong answer is still in equilibrium with the same loads. Equilibrium is one equation per degree of freedom of the whole body, and a stiffness error lives entirely in the many equations underneath it. A model can satisfy every equilibrium check ever devised and be a model of a different structure. Equilibrium

The check that cannot see the error

Every analysis prints a global equilibrium residual, and it is the first thing anybody looks at. It catches a lost restraint and a load entered in the wrong unit immediately. It is structurally incapable of catching a member whose stiffness is wrong by a factor of ten, because the wrong answer is still in equilibrium with the same loads.

 The gap is a sum of five things and only one of them is computed. What a 30 mm movement joint is asked to accommodate, by three combination rules. The top bar is every term at its extreme, added: 33.2 mm, which assumes the hottest day, the fullest floor, the whole of the shrinkage and the worst-placed wall arrive together. The chance of that is about 1.5%. The bottom bar treats them as independent and asks for 16.0 mm. The middle bar is the rule used for actions and almost never for movements — one term at its full value and the rest at their coincidence factors — and gives 25.5 mm. The segments across the top bar are the terms themselves, and the ordering is the finding: the largest is tolerance at 10.0 mm, which is not a structural quantity at all, and the smallest is deflection at 3.2 mm — the only one anybody computes carefully, and 10% of the total. Deflection

The gap nobody computed

A movement joint is sized by adding up everything the structure will do to it, and the deflection calculation — the only term anybody computes carefully — is usually the smallest one in the list. The largest is a construction tolerance, which is not a structural quantity at all, and the sum of the extremes is nearly twice what treating them as independent would ask for.

 The strength was bought in a furnace and the welder gives it back. Proof stress as delivered and beside a weld, for four aluminium alloys. The heat-treated alloys lose half of it: the strength of a 6xxx extrusion is in precipitates formed by an ageing treatment, and the arc dissolves them for 32 mm either side of the weld, permanently. The work-hardened tempers lose nearly as much, because the heat undoes exactly the work. The annealed ones lose nothing at all, because there is nothing left in them to anneal. The consequence is the crossover: 5083-H22 is 1.04 times 6061-T6 as delivered and 0.92 times it once welded, so the stronger alloy is the weaker member. The 240 mm member drawn, with two longitudinal welds, keeps 87% of its parent capacity — a weld along a member softens a strip and leaves a section, and the same weld across it softens the whole of one. Materials

The strength the welder gives back

A 6082-T6 extrusion is twice as strong as a 5083-H111 plate and, welded across, the two are within a few per cent of each other. The heat of the arc anneals the metal for thirty millimetres either side, permanently, and the strength that was bought in a furnace is given back at the first joint.

 The bearing that is drawn as a roller. The horizontal force a sliding bearing delivers, against the vertical load it is carrying, with its coefficient of friction on the same picture. The coefficient is not a constant: PTFE's falls as the contact pressure rises, and the standard fit is μ = 1.2/(10 + σ), so the bearing drawn is at 30.0 N/mm² and μ = 0.030 while the same bearing at a fifth of the load is at 0.075 — 2.5 times as much. The force curve is therefore strongly non-linear: a fifth of the load gives 50% of the force. Two readings follow and only one of them is usually taken. The largest force is at full load, 108 kN, and that is what the pier is designed for. The largest nuisance is at light load, where 54 kN of friction is 39% of the 140 kN of wind the bearing was put there to release the structure from. Cold makes it worse again: below about −5 °C the same bearing delivers 216 kN. A roller symbol on a drawing means this, and it is a pair of load cases rather than one, because friction opposes whichever way the deck happens to be going. Connections

The roller that is not a roller

A sliding bearing is drawn as a roller and detailed as a sheet of PTFE, and it delivers a horizontal force of a few per cent of whatever it is carrying. The coefficient everybody quotes is the one at full design pressure, and PTFE's coefficient rises as the pressure falls — so the bearing is at its freest exactly where nobody checks it.

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